A true-false test consists of 15 items.
(b) What is the probability that Chris gets 80% or more for the test?
(c) If it is a 25 item true-false test, would you think it is easier or more difficulty for Chris to get 80% or more? (Hint: please calculate the corresponding probability that Chris gets 80% or more for the test, and compare your results with the one obtained in (b) to answer this question.)
Answer.
b) he must get 13 right or 14 right or 15 right
= C(15,13)(1/2)^13 (1/2)^2 + C(15,14)(1/2)^14(1/2) + C(15,15) (1/2)^15
= 0.01922%
c) he must get 23 right or 24 right or 25 right
= C(25,23)(1/2)^23 (1/2)^2 + C(25,24)(1/2)^24(1/2) + C(25,25) (1/2)^25
= 5.36441803e-0.5
The assumption would be that he would be randomly guessing at the answer.
Since 12/15 = .8, why did you not include the probability of getting 12 correct out of the 15 ?
Try adding that to your other cases.
For b) wouldn't getting 20 out of 25 correct yield 80% ?
Same for 21 and 22
So you have a few more cases to include
maybe speak to your tutor if you're having issues. very easy to get done for plagiarism for this when it shows up as the first result on google. QUT is coming for you
To calculate the probability in each case, we need to use the binomial probability formula. The formula for calculating the probability of k successes in n independent Bernoulli trials, where each trial has a probability p of success, is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- C(n, k) represents the number of combinations of n items taken k at a time (also known as binomial coefficient)
- p is the probability of success
- k is the number of desired successes
- n is the total number of trials
Now let's calculate the probabilities step by step:
b) In a 15-item test, Chris must get 13, 14, or 15 items correct to achieve a score of 80% or more. The probability of getting a single item correct is 1/2 since it's a true-false test. Therefore, p = 1/2.
P(X ≥ 13) = P(X = 13) + P(X = 14) + P(X = 15)
= C(15, 13) * (1/2)^13 * (1/2)^2 + C(15, 14) * (1/2)^14 * (1/2) + C(15, 15) * (1/2)^15
= 0.01922 (rounded to 5 decimal places)
So, the probability that Chris gets 80% or more in a 15-item true-false test is approximately 0.01922, or 1.922%.
c) In a 25-item test, Chris must get 23, 24, or 25 items correct to achieve a score of 80% or more. Using the same probability of success (p = 1/2) as before:
P(X ≥ 23) = P(X = 23) + P(X = 24) + P(X = 25)
= C(25, 23) * (1/2)^23 * (1/2)^2 + C(25, 24) * (1/2)^24 * (1/2) + C(25, 25) * (1/2)^25
= 5.36441803 * 10^(-5) (rounded to 8 decimal places)
So, the probability that Chris gets 80% or more in a 25-item true-false test is approximately 5.36441803e-5, or 0.0000536441803, which is much smaller than the probability in part (b) for a 15-item test.
Therefore, based on the calculations, we can conclude that it would be more difficult for Chris to get 80% or more in a 25-item test compared to a 15-item test.